Physics - Gravitation Question with Solution | TestHub

Kinetic energy of satellite$\left(E\right)$

$KE\left(k\right)=\frac{1}{2}m{v}^2$

$=\frac{1}{2}m{\left(\sqrt{\frac{GM}{R}}\right)}^2$$\left(\because v=\sqrt{\frac{GM}{R}}\right)$

[$m=$mass of satellite]

$=\frac{1}{2}m\frac{GM}{R}$

[Symbols have usual meanings]

Potential energy of satellite,

$U=-m\frac{GM}{R}$

$\therefore$Binding energy of the satellite,

$E_b=KE+U$

$=\frac{1}{2}m\frac{GM}{R}+\left(-m\frac{GM}{R}\right)$

$=\frac{1}{2}m\frac{GM}{R}-m\frac{GM}{R}$

$=-\frac{GMm}{2R}$

$\therefore$Magnitude of the binding energy$E=\left|E_b\right|=\frac{GMm}{2R}$

The ratio of$\frac{E}{K}=\frac{GMm}{2R}\times \frac{2R}{GMm}=1$